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Lesson Explainer: Relating Volumes and Surface Areas Mathematics • 8th Grade

In this explainer, we will learn how to relate volumes of different shapes, surface areas of different shapes, and a shape’s volume and surface area.

Since we are interested in relating the volumes and surface areas of shapes, let us begin by recalling the formulas that exist for calculating these quantities. First of all, let us consider probably the simplest case, spheres.

Formula: Volumes and Surface Areas of Spheres

Let π‘Ÿ be the radius of a sphere. Then, its volume 𝑉 is given by 𝑉=43πœ‹π‘Ÿ, and the surface area 𝑆 is given by 𝑆=4πœ‹π‘Ÿ.

Note that both formulas only have one variable quantity: the radius π‘Ÿ. This means that, if we are given the radius, we can find both the volume and the surface area. However, that is not all. Suppose we are given the surface area of a sphere and want to find other measurements of the sphere. Since the surface area is 𝑆=4πœ‹π‘Ÿ, we can find the radius π‘Ÿ by rearranging the formula in terms of it. We first divide both sides by 4πœ‹ to get 𝑆4πœ‹=π‘Ÿ.

Then, as π‘Ÿ is always positive, we can take the square root of both sides: ο„žπ‘†4πœ‹=π‘Ÿ.

So, from the surface area, we can get the radius. Furthermore, since the volume is 𝑉=43πœ‹π‘Ÿ, we can substitute π‘Ÿ directly into this equation to find the volume as well.

This process also applies the other way around. If we are given the volume, we can find the radius and, by extension, the surface area. Let us see an example of this.

Example 1: Relating the Volume and Surface Area of a Sphere

Given that the volume of a sphere is 562.5πœ‹ cm3, find its surface area in terms of πœ‹.

Answer

We begin by considering the information given to us and what relevant formulas we can apply. We have been given the volume (𝑉) of a sphere, which we can recall is given by 𝑉=43πœ‹π‘Ÿ, where π‘Ÿ is the radius of the sphere. We have also been asked to find the surface area (𝑆), which is given by 𝑆=4πœ‹π‘Ÿ.

Since both formulas involve π‘Ÿ, we can relate the two. Specifically, we can use the volume formula to find the radius, which we can then substitute into the surface area formula to find 𝑆. We start by substituting 𝑉=562.5πœ‹ into the volume formula to get 562.5πœ‹=43πœ‹π‘Ÿ.

Dividing both sides by 43πœ‹, we have 562.5Γ—34=π‘Ÿ.

Then, to find π‘Ÿ, we can take the cube root of both sides. This gives us π‘Ÿ=ο„ž562.5Γ—34=7.5.cm

Now, we can substitute π‘Ÿ=7.5cm into the surface area formula. This gives us 𝑆=4πœ‹Γ—7.5=225πœ‹.cm

Since we have been asked to find the surface area in terms of πœ‹, we can leave the answer as it is without simplifying.

As we saw in the last example, the volume and surface area of a sphere are closely related to each other. This also applies to other kinds of shapes; for instance, let us consider cylinders. First, let us recall the formulas for the volume and surface area of a cylinder.

Formula: Volumes and Surface Areas of Cylinders

For a cylinder, let π‘Ÿ be the radius of the base and let β„Ž be the height. Then, its volume 𝑉 is given by 𝑉=πœ‹π‘Ÿβ„Ž, and its surface area 𝑆 is given by 𝑆=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ.

Recall that both of these formulas come from the formulas for the area and circumference of a circle. Firstly, the volume is the area of the circular base 𝐴=πœ‹π‘Ÿο…οŠ¨ times the height β„Ž, which is shown in the left-hand diagram below. For the surface area, we consider the net of the cylinder, which is the area of two circles ο€Ή2𝐴=2πœ‹π‘Ÿο…οŠ¨, plus the curved area, which is the circumference (𝐢=2πœ‹π‘Ÿ) times the height β„Ž. This is shown in the right-hand diagram below.

Considering the above formulas, we see that they are both in terms of two variables: the radius π‘Ÿ and the height β„Ž. We highlight this similarity below.

In other words, as long as we have the radius and height, it is possible to find both the volume and surface area. Additionally, if we are given either the radius or the height and either the volume or the surface area, it is possible to find the other two measurements. This is because we can rearrange the formulas to be in terms of the missing variables.

To show how this works, let us consider an example where we are given the curved surface area and the diameter of a cylinder and need to find the volume.

Example 2: Relating the Volume and Surface Area of a Cylinder

A cylinder has a curved surface area of 656 cm2 and a diameter of 10.2 cm. Find its volume, giving your answer to the nearest cubic centimetre.

Answer

Let us first of all understand the information given to us. We have been given that the curved surface area is 656 cm2, which is a portion of the total surface area, as shown below.

In other words, the curved surface area is the area of the rectangle shown, which is 2πœ‹π‘Ÿ times β„Ž. Additionally, we have been given that the diameter is 10.2 cm, where the diameter is twice the radius π‘Ÿ. Since we have been given two measurements of the cylinder, we can use them to find the height, and consequently, the volume.

First, for the curved surface area, we have 656=2πœ‹π‘Ÿβ„Ž.

Using the fact that 2π‘Ÿ=𝑑=10.2cm, where 𝑑 is the diameter, we have 656=10.2πœ‹β„Ž.

To find β„Ž, we divide both sides by 10.2πœ‹ to get β„Ž=65610.2πœ‹.cm

Now, we can use this to find the volume. Recall that the volume 𝑉 of a cylinder is 𝑉=πœ‹π‘Ÿβ„Ž.

We substitute in π‘Ÿ=10.22=5.1cm and β„Ž=65610.2πœ‹cm to get 𝑉=πœ‹Γ—5.1Γ—65610.2πœ‹=5.1Γ—65610.2=1673.cmnearestcm

One other type of shape we will commonly see is cuboids (and cubes). Let us recall the formulas for the volume and surface area of a cuboid.

Formula: Volume and Surface Area of Cuboid

Suppose a cuboid has length 𝑙, width 𝑀, and height β„Ž, as shown in the figure below. Then, its volume 𝑉 is given by 𝑉=π‘™π‘€β„Ž, and its surface area 𝑆 is given by 𝑆=2(𝑙𝑀+π‘€β„Ž+β„Žπ‘™).

The formula for the volume is fairly straightforward, since it is just the product of the three dimensions of the cuboid. For the surface area, recall that this can be found by considering the net of the cuboid, as shown below.

In other words, the areas of each of the 6 faces are added together, where the area of each face is the product of two dimensions of the cuboid.

We also note that the above formulas easily extend to cubes, since a cube is just a cuboid with equal sides. In this case, 𝑙=𝑀=β„Ž, so the volume becomes 𝑉=𝑙, and the surface area becomes 𝑆=6𝑙.

So far, we have seen examples of relating the surface area to the volume of individual shapes, but it is equally possible to relate the volumes and surface areas of different shapes together. Let us consider a case where a sphere is contained within a cube.

Example 3: Relating the Volumes of a Cube and a Sphere

If a sphere is inscribed in a cube of volume 8 cm3, what is the volume of the sphere?

Answer

The key to this question is being able to relate the dimensions of the cube to the sphere. We note that the sphere is inscribed in the cube, which means that the sphere is touching each face of the cube without any gaps. This means that the diameter 𝑑 of the sphere is equal to the length of the cube (or alternatively, its radius π‘Ÿ is half that of the length of the cube). This is demonstrated below.

Therefore, if we can find the length of the cube, then we can find the diameter (and radius) of the sphere.

Now, we have been given that the volume of the cube is 8 cm3. Recall that if the length of a cube is 𝑙, then its volume π‘‰οŠ§ is given by 𝑉=𝑙.

Substituting 𝑉=8, we have 8=𝑙.

We can obtain 𝑙 by taking the cube root of both sides of the equation. This gives us 𝑙=2.cm

As mentioned earlier, the radius π‘Ÿ is half this length. So, this means that π‘Ÿ=1.cm

Now that we have the radius of the sphere, we can obtain its volume π‘‰οŠ¨ by using the formula 𝑉=43πœ‹π‘Ÿ.

Since π‘Ÿ=1, we have 𝑉=43πœ‹Γ—1=43πœ‹.cm

As demonstrated in the previous example, when comparing different objects together, it is often important to recognize when their dimensions such as the length, height, or radius are related to each other. We can do this using information given to us in the problem and deductive reasoning. Let us see another example of relating two objects together.

Example 4: Relating the Surface Areas of a Cuboid and a Cylinder

If the total surface of a rectangular prism is 40 cm2 and its height is twice its width, find the total surface area of a cylinder inscribed inside the rectangular prism.

Answer

Let us begin by analyzing the information given in the question. The cylinder has been inscribed within a cuboid, which means that the edges of the two shapes are touching. Let us consider what this means by looking at a cross-section (i.e., a top-down view) of the two shapes.

Here, we have labeled the two diameters of the cylinder 𝑑, the cuboid’s width 𝑀, and the cuboid’s length 𝑙. We can see from the diagram that the width and length must each be equal to the diameters, which means 𝑀=𝑙.

Secondly, from the question, we have been given that the height β„Ž is twice the width (i.e., β„Ž=2𝑀). This means that all three dimensions of the cuboid can be expressed in terms of the width. We label this on the diagram below.

Here, π‘Ÿ is the radius of the cylinder, and we note that the height of the cylinder is the same as the height of the cuboid, 2𝑀.

Let us now apply the formulas we know for the surface areas of cylinders and cuboids. Ordinarily, the surface area π‘†οŠ§ of a cuboid is given by 𝑆=2(𝑙𝑀+π‘€β„Ž+β„Žπ‘™), but here, we have 𝑙=𝑀 and β„Ž=2𝑀. We have also been given that the total surface area is 40 cm2. Substituting these values, we get 40=2(𝑀×𝑀+𝑀×2𝑀+2𝑀×𝑀)=2𝑀+2𝑀+2𝑀=10𝑀.

From this, we can find the width. We first divide by 10 on both sides to get 4=𝑀, and, since 𝑀 is a length (i.e., it must always be positive) we can take the square root of both sides as follows: 𝑀=2.cm

Now, recall that the radius π‘Ÿ of the cylinder is half of 𝑀, so π‘Ÿ=1cm. We now recall that the total surface area π‘†οŠ¨ of a cylinder is the area of the top and bottom circles plus the circular portion, which is 𝑆=2πœ‹π‘Ÿ+2πœ‹π‘Ÿβ„Ž.

Here, we have π‘Ÿ=1 cm and β„Ž=2𝑀=4cm, so we substitute these values into the formula to get 𝑆=2πœ‹Γ—1+2πœ‹Γ—1Γ—4=2πœ‹+8πœ‹=10πœ‹.cm

For our final example, we will consider a situation where multiple spheres are combined together and we have to consider their relation to a cylinder.

Example 5: Relating the Volumes and Surface Areas of Multiple Shapes

Three spheres are inscribed in a cylinder, as shown in the figure. If the volume of a sphere is 36πœ‹ cm3, find the total surface area of the cylinder.

Answer

In this question, we need to consider how a single sphere relates to the dimensions of the cylinder that contains the three spheres. To do this, we can make use of the fact that the spheres are inscribed, which means they fit exactly into the cylinder with no gaps in the top, bottom, or sides. In particular, this means the radii of the spheres and the cylinder must be the same. It also means that the height of the cylinder must be equal to 3 times the diameter of a single sphere (or 6 times the radius of one). Letting π‘Ÿ be the radius, this is illustrated below.

To find the surface area of the cylinder, it will be necessary to find π‘Ÿ since that will give us the radius and height of the cylinder.

Since we have been given the volume of a sphere, let us recall that the volume 𝑉 is related to π‘Ÿ by the formula 𝑉=43πœ‹π‘Ÿ.

Letting 𝑉=36πœ‹cm, we have 36πœ‹=43πœ‹π‘Ÿ.

To get π‘Ÿ from here, we can rearrange the equation by dividing both sides by 43πœ‹. This gives us π‘Ÿ=36πœ‹Γ·4πœ‹3=27.

Next, we can find π‘Ÿ by taking the cube root of both sides. This gives us π‘Ÿ=3.cm

Now, let us recall the formula for the surface area 𝑆 of a cylinder. This is equal to the area of the two circles forming the top and bottom plus the area of the curved region, which is equal to 𝑆=2πœ‹π‘Ÿ+2πœ‹π‘Ÿβ„Ž.

We have already shown that π‘Ÿ=3cm for the cylinder since it has the same radius as one of the spheres, and the height is 6π‘Ÿ=18cm. Substituting these values in, we get 𝑆=2πœ‹Γ—3+2πœ‹Γ—3Γ—18=18πœ‹+108πœ‹=126πœ‹.cm

Let us finish by considering the main things we have learned in this explainer.

Key Points

  • For a sphere, if we have any one of the volume 𝑉, radius π‘Ÿ, or surface area 𝑆, we can find the other two measurements using the formulas 𝑉=43πœ‹π‘Ÿ,𝑆=4πœ‹π‘Ÿ.
  • For a cylinder, if we have the radius π‘Ÿ and height β„Ž, then we can find the volume 𝑉 and surface area 𝑆 using the formulas 𝑉=πœ‹π‘Ÿβ„Ž,𝑆=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ. Furthermore, if we are given either the volume or the surface area and either the radius or the height, we can calculate the other two measurements by rearrangement.
  • For a cuboid (or cube), if we have the length 𝑙, width 𝑀, and height β„Ž, then we can find the volume 𝑉 and surface area 𝑆 using the formulas 𝑉=π‘™π‘€β„Ž,𝑆=2(𝑙𝑀+π‘€β„Ž+β„Žπ‘™). Additionally, if we are given two of the dimensions of a cuboid and either the volume or surface area, we can find the remaining measurements by rearrangement.
  • In problems where we are given different shapes, we can relate them to each other by using the information given to us in the problem and the rearrangement of the above formulas.

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